Question

You place a 500 g block of an unknown substance in an insulated container filled 2 kg of water. The block has an initial temperature of 50 degrees C. The water is initially at 20 degrees C. If the equilibrium temperature of the block and water is 25 degrees C, what is the specific heat of the block? The specific heat of water is 4186 J. kg K. 3349 J, kg C 1189 J, kg C 5545 J, kg C 750 J/kg C 2080 J/kg C

Answer 1

Answer:

3349J/kgC

Explanation:

Questions like these are properly handled having this fact in mind;

• Heat loss = Heat gained

Quantity of heat = mcΔ∅

m = mass of subatance

c = specific heat capacity

Δ∅ = change in temperature

m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)

m₁ = mass of block = 500g = 0.5kg

c₁  = specific heat capacity of unknown substance

∅₂ = block initial temperature = 50oC

∅₁ = equilibrium temperature of block and water after mix= 25oC

m₂= mass of water = 2kg

c₂ = specific heat capacity of water = 4186J/kg C

∅₃ = intial temperature of water = 20oC

0.5c₁(50-25) = 2 x 4186(25-20)

And we can find c₁ which is the unknown specific heat capacity

c₁ = $\frac{2*4186*5}{0.5*25}$= 3348.8J/kg C≅ 3349J/kg C